1
Monomial Orders.
In the polynomial algebra over F a …eld in one variable x; F [x], we can do
long division (sometimes incorrectly called the Euclidean algorithm). If f (x) =
a0 + a1 x + ::: + an xn 2 F [x] with an 6= 0 then we write deg f (x) = n and
LC(f (x)) = am . If g(x) is another element of F [x] then we have
f (x) = h(x)g(x) + r(x)
with h(x); r(x) 2 F [x] and deg r[x] < m. This expression is unique. This result
can be vari…ed by long division. As with all divisions we assume g(x) 6= 0.
Which we will recall as a pseudo code (such code terminates on a return) the
input being f; g and the output h and r:
f0 = f ; h0 = 0; k = 0; m = deg g;
Repeat:
If deg(fk (x) < deg(g(x) return hk (x); fk (x);n = deg fk;
k (x)) n m
g(x)
fk+1 (x) = fk (x) LC(f
LC(g(x)) x
k (x)) n m
hk+1 (x) = hk (x) + LC(f
;
LC(g(x)) x
k = k + 1;
Continue;
We note that this code terminates since at each step when there is no return
then the new value of the degree of fx (x) has strictly decreased.
The theory of Gröbner bases is based on a generalization of this algorithm
to more than one variable. Unfortunately there is an immediate di¢ culty. The
degree of a polynomial does not determine the highest degree part of the polynomial up to scalar multiple. But this is precisely what is needed. To …x this
problem one introduces what is known as a monomial order.
We say that a total order (i.e satis…es transitivity and tricotomy), >, on
Zn 0 is a monomial order if it satis…es the following conditions (Greek letters
will designate elements of Zn 0 ):
1. 6= (0; :::; 0) implies that > 0.
2. If > then + > + .
3. Any non-empty subset of Zn 0 has a minimal element.
Exercise 1. Show that the only possible order on Z 0 satsifying 1. and 2.
is the usual order.
Exercise 2. Show that condition 3 is equivalent with the condition: If
then there exists M such that k = M for
1
2
3
m
k M.
We will now explain two of the main ways of getting monomial orders on
q
p
Zp+q
0 if we have monomial orders >1 on Z 0 and >2 on Z 0 .
I. Write = ( ; ); = ( ; ) with ; 2 Zp 0 and ; 2 Zq 0 then the new
order is given as follows: > if >1 or if = and >2 :
If 6= 0 then if = ( ; ) if 6= 0 then >1 0 so > 0 if = 0 then if 6= 0
then 6= 0 hence >2 0 hence > 0. If > = ( ; ) then writing = ( ; )
we see that if >1 then + >1 + if = then + = + and >2
which implies + >2 + . Thus condition 2 is satsi…ed. Condition 3 is
1
satsi…ed since in such a sequence the …rst components are in decreasing order
thus must stabilize. Once they stabilize the second omponents are in decreasing
order.
II. Here there are three levels of tests. Let = ( ; ) and = ( ; ). If
j j > j j then > if j j = j j and <2 then > (notice the reversal
here) If j j = j j and = and if >2 then > .
We leave it to the reader to check that this de…nes a monomial order.
If we start with the only possible order on Z 0 then use method I on Z2 0
then method I. on Z3 0 = Z 0 Z2 0 and continue in this way to Zn 0 then we
get Lexicographic order or Lex order. That is, if
= (0; :::; 0; u; :::) and
u > 0 then > . We write >Lex for this order.
The other standard order is Graded Reverse Lex or GrRevLex. Which is
gotten by an iteration of method II. Here the …rst test is on the norm. Next
we look at the last entry. If the last entry of is strictly less than that of
then > otherwise if one has equality and n > 2 we use the same test on the
second to last entry (if n = 2 and then at this stage we know = ). If one
has equality in the next to last index and n > 3 then one goes to the next index
up and uses the same test, etc. We write >GrRLex for this order.
Exercise 3. If > is a total order on Zn 0 that satis…es 2. in the de…nition of
monomial order. Then we introduce a new version of the order by saying that
if j j > j j then >Gr and if j j = j j and if > then >Gr . Show that
>Gr de…nes a monomial order.
The order in exercise 3 is called the graded version of >. We denote by
GrLex the graded version of Lex and denote the order by >GrLex .
Exercise 4. Show that if n = 2 then GrLex is the same as GrRevLex.
If x and x are monomials and if > is a monomial order we will write
x > x if > .
Fix a monomial order, >. (We will use it throught the rest of this section).
If f (x1 ; :::; xn ) 6= 0 is a polynomial we can write
X
f (x1 ; :::; xn ) = a x +
a x :
<
We will call x the leading monomial of f (x1 ; :::; xn ) and denote it by LM (f (x1 ; :::; xn ))
(we will also write = LM (f (x1 ; :::; xn )). We will call a the leading coe¢ cient
of f (x1 ; :::; xn ) and denote it LC(f (x1 ; :::; xn ).
If ; 2 Zn 0 then we say that
dominates
if
2 Zn 0 . With
all of this in place we can set up the multivariate division algorithm. Here
f (x1 ; :::; xn ); g(x1 ; :::; xn ) 2 F [x1 ; :::; xn ].
h0 (x1 ; :::; xn ) = 0; f0 (x1 ; :::; xn ) = f (x1 ; :::; xn ); = LM (g(x)); k = 0;
Repeat:
If LM (fk (x)) does not dominate return hk (x1 ; :::; xn ); fk (x1 ; :::; xn );
k (x1 ;:::;xn )) LM (fk (x1 ;:::;xn )
g(x)
fk+1 (x1 ; :::; xn ) = fk (x1 ; :::; xn ) LC(f
LC(g(x1 ;:::;xn )) x
k (x1 ;:::;xn )) LM (fk (x1 ;:::;xn )
hk+1 (x) = hk (x) + LC(f
LC(g(x1 ;:::;xn )) x
k = k + 1;Continue;
2
;
The output of this algorithm is a pair of polynomials h(x1 ; :::; xn ) and
r(x1 ; :::; xn ) with
f (x1 ; :::; xn ) = h(x1 ; :::; xn )g(x1 ; :::; xn ) + r(x1 ; :::; xn )
and the leading monomial of r(x1 ; :::; xn ) does not dominate that of g(x1 ; :::; xn ).
We will call r(x1 ; :::; xn ) the reduction of f (x1 ; :::; xn ) relative to g(x1 ; :::; xn )
relative to the given monomial order. We will denote r be remainder(f; g).
Exercises 5. Calculate h and r for f = x4 y 4 + x2 y 6 and g = x3 y + x2 y 3
using Lex and GrLex.
6. Show that if f = ug then the division algorthm outputs h = u and r = 0.
If we have a set of nonzero polynomials f1 ; :::; fm than will now describe
a procedure analogous to Gaussian elimination to put them in a convenient
“normal form”. In which we relace the polynomials by polynomials h1 ; :::; hl
generating the same ideal with the following properties:
1. If i 6= j then LM (hi ) does not domiante LM (hj ):
2. LM (h1 ) < LM (h2 ) < ::: < LM (hl ).
We now give the normalize algorithm. Here we partially order polynomials
by their leading monomials using the monomial order that we have …xed. The
statement Break means that if one is doing operations 1; 2; ::: and the Break is
at i then one stops doing that operation immediately (it takes the ith step as
the last in the loop).
For i = 1; :::; m fi;0 = fi ; k = m; l = 0;
Repeat:
Remove all of the fi;l that are 0. Relabel the rest so that they are in the
same order as before. Subtract the number of zeros deleted from k.
Order the f1;l ; :::; fk;l in the …rst subindex (remember the partial order is
according to the order on the lead monomials).
T ST = ( 1; 1);
For i = 1; i
k 1,If T ST 6= ( 1; 1) Break; For j = i + 1; j
k if fj;l
dominates fi;l then T ST = (i; j) = (T ST1 ; T ST2 );Break;
If T ST = ( 1; 1) then Return f1;l ; :::; fk;l ;
For s = 1; s < T ST2 ; fs;l+1 = fs;l ; fT ST2 ;l+1 = remainder(fT ST2 ;l ; fT ST1 ;l );
For s = T ST2 + 1; s k; fs;l+1 = fs;l ;
l = l + 1;
Continue.
We must check that this algorithm terminates. Suppose not then at each
stage there must be a pair of indices such that a nontrivial division has taken
place. That is at each stage l there is some i such that LM (fi;l ) > LM (fi;l+1 ):
Since there are only m possible values of i there must be an in…nite sequence
l1 < l2 ; ::: such that LM (fi;lj ) > LM (fi:lj+1 ). This contradicts 3. in the
de…nition of monomial order.
We will now give the reduction algorithm that will be important to our
development of Gröbner bases. Let G = fg1 ; :::; gm g be a set of non-zero poly3
nomialsin a given order and let f be a polynomial. Here is the algorithm (see
the previous algorithm for the meaning of Break).
f0 = f ; a1 = a2 = ::: = am = 0; k = 0;
Repeat:
T ST = 1;
For i = 1 to m if LM (fk ) dominates LM (gi ) then T ST = i and Break;
If T ST == 1 return (a1 ; :::; am ); fk ;
Do division algorithm on fk and gT ST with output b; fk+1 ;
ai = ai + b;
k = k + 1;
Continue;
The output of this algorithm is (a1 ; :::; am ); r: With r; ai 2 F [x1 ; :::; xn ],
f = a1 g1 + ::: + am gm + r and LM (r) does not dominate any of the LM (gi ).
The polynomial r will be called the reduction of f with respect to the ordered
set g1 ; :::; gm .
If LM (gi ) 6= LM (gj ) for i 6= j and LM (gi ) < LM (gi+1 ) for i = 1; :::; m 1
then we will use the notation redG (f ) for the output r of the above algorithm.
Exercise 7. Show that this algorithm terminates.
Exercise 8. Suppose that in the above algorithm we used a di¤erent ordering of the elements of G would we get the same r?
2
Gröbner bases.
If I is an ideal in F [x1 ; :::; xn ] and if > is a monomial order then we de…ne
LM (I) to be the ideal generated by fLM (f )jf 6= 0; f 2 Ig. We call LM (I) the
monomial ideal associated with I and the order >.
If I is an ideal and if > is a monomial order then we set S(I) equal to the
set of monomials that do not dominate any element of fLM (f )jf 6= 0; f 2 Ig.
L
Lemma 1 F [x1 ; :::; xn ] = I
SpanF S(I). Furthermore,S(I) is a linearly independent set.
Proof. The second assertion is obvious. That the sum in the asertion is direct is
also obvious (we leave this as an L
exercise with the hint that it is obvious). Let A
be the set of all f (x1 ; :::; xn ) 2
=I
SpanF S(I). We assume that A is nonempty.
Then the set fLM (f )jf 2 Ag has a minimal element. Let f 2 A be such that
LM (f ) is that element. We now assume that LC(f ) = 1: If LM (f ) 2
= S(I) then
LM (f ) dominates LM (g) for some g 2 I. We may assume that LC(g) = 1.
LM (f )
(f )
Then LM (f LM
LM (g) g) < LM (f ) and f
LM (g) g 2 A. This is a contradiction.
If LM (f ) 2 S(I) then LM (f LM (f )) < LM (f ) and we run into the same
contradiction.
Corollary 2 The dimension of the variety de…ned by LM (I) is the same as
that de…ned by I.
4
Proof. The value of the Hilbert function, h(k); of F [x1 ; :::; xn ]=I equal to the
number of elements, , of S(I) with j j
k. This is the same value as for
F [x1 ; :::; xn ]=LM (I).
One can see that the monomial ideal of I relative to the monomial order > is
an important computational tool. Our problem is to …nd a way to calculate it.
We will …rst introduce another concept that is also apparently non-algorithmic
and will (following Buchberger) …nd an algorithm to calculate it.
Lemma 3 Let > be a momomial order and let I be an ideal in F [x1 ; :::; xn ].
Let fg1 ; :::; gm g be a set elements of I. Then the following are equivalent:
1. The set fLM (g1 ); :::; LM (gm )g generates LM (I).
2.Order the gi arbitrarily. If f is a polynomial then f 2 I if and only if the
reduction of f with respect to g1 ; :::; gm is 0.
We will call a set of generators for I that satis…es the conditions of the
lemma for > a Gröbner basis for I with respect to >. We note that 1. implies
that Gröbner bases always exist since we can apply the Hilbert basis theorem to
LM (I) to get a …nite generating set x 1 ; :::; x m for LM and choose g1 ; :::; gm
in I so that LM (gi ) = x i . We also note that 2. implies that a Gröbner basis
is a generating set of I.
We will now prove the result.
Assume that fg1 ; :::; gm g satis…es 1. Then if f 2 I we have LM (f ) must
dominate some element of fLM (g1 ); :::; LM (gm ). Let LM (gi ) be the one with
smallest index. If we apply the division algorithm to f and gi the remainder has
a lead monomial strictly less than that of f . So let A be the set of all f such
that 2. is not satis…ed. Let f be an element of A with LM (f ) minimal. Now
use the procedure just discribed derive a contradiction. Clearly if f reduces to
0 by repeated applications of the division algorithm then f 2 I.
We will now show that 2. implies 1. We must only show that if f 2 I;
f 6= 0 then LM (f ) dominates LM (gi ) for some i. But this is obvious since it is
necessary in order to begin the reduction of f to 0 via the division algorithm.
As an immediate corollary we have:
Corollary 4 Let G = fg1 ; :::; gm g be such that LM (gi 6= LM (gj ) for i 6= j.
Then G is a Gröbner basis for hGi if and only if f 2 F [x1 ; :::; xn ] is in hGi if
and only if redG (f ) = 0.
Notice that if we have a Gröbner basis of I with respect to > then we can
test whether f 2 I using the reduction algorthm. In fact, the algorithm shows
us how to write f in terms of the gi . However much more is true.
We say that a Gröbner basis, fg1 ; :::; gm g for I relative to > is reduced if no
monomial of gi dominates LM (gj ) for j 6= i.
Lemma 5 Given a monomial order > each ideal has a unique reduced Gröbner
basis with leading all coe¢ ecients 1..
5
Proof. Let fg1 ; :::; gm g be a Gröbner basis for I with respect to the monomial
order >. We may assume that all of the leading coe¢ cients of the gi are 1. We
note that if LM (gi ) dominates LM (gj ) for j 6= i then fg1 ; :::; gm g fgi g is still
a Gröbner basis. We can thus assume that no LM (gi ) dominates LM (gj ) for
i 6= j. We can now assume that this property is satis…ed and that LM (g1 ) <
LM (g2 ) < ::: < LM (gm ). We note that if x is a monomial with non-zero
coe¢ cient in gi then
LM (gi )thus cannot dominate LM (gj ) for j > i.
Thus no monomial of g1 dominates LM (gi ) for i > 1. We now consider the
non-leading monomials of g2 . Let
X
gi =
a ;i x
Let
2
= f ja
;2
6= 0;
dominates LM (g1 )g replace g2 by
X
g2
a ;2 x LM (g1 ) g:
2
2
Now no monomial of g2 dominates the lead monomial of gi for i 6= 2. We now
consider g3 we …rst do exactly the above procedure to g3 relative to g2 so that
the new g3 has no monomial that dominates LM (g2 ). Now do the same with
g3 and g1 : We next do the same thing with g4 relative to gi ; i = 1; 2; 3. When
we get to m we have a reduced Gröbner basis. This shows that from a Gröbner
basis we can always algorithmically construct a reduced Gröbner basis.
Let fg1 ; :::; gm g and fh1 ; :::; hl g be reduced Gröbner bases of I with respect
to > with leading coe¢ ecients 1.. We …rst assert that fLM (g1 ); :::; LM (gm )g =
fLM (h1 ); :::; LM (hl )g. Indeed LM (gi ) must dominate LM (hj ) for some j and
LM (hj ) must dominate LM (gk ) for some k. Thus LM (gi ) dominates LM (gk )
so we must have k = i. This implies LM (gi ) = LM (hj ). This proves the
assertion. Thus m = l and we can reorder so that LM (gi ) = LM (hi ). We note
that if gi hi 6= 0 then LM (gi hi ) < LM (gi ): But LM (gi hi ) must dominate
some LM (gj ) this implies that either gi or hi as a non-leading monomial that
domiates one of the LM (gj ). This contradicts the de…nition of reduced.
This result says that if we can e¤ectively construct Gröbner bases then we
have an e¤ective test for whether two ideals are equal.
Exercises 1. Show that the procedure described in the …rst part of the
proof does produce a reduced Gröbner basis.
2. Write pseudocode to convert a Gröbner basis to a reduced Gröbner basis.
3
Syzygies.
In this section we will give a test (due to Buchberger) as to when a generating
set for an ideal is a Gröbner basis relative to a monomial order. Through this
section we keep a …xed monomial order >. If f; g 6= 0 are polynomials then
let x be the least common multiple of LM (f ) and LM (g): We note that if
6
LM (f ) = x and LM (g) = x then
=
and =
then we set
=(
S(f; g) = LC(g)x f
1 ; :::;
n)
with
i
= max( i ;
i ).
If
LC(f )x g:
Buchberger’s criterion is:
Theorem 6 Let I be an ideal. Let G = fg1 ; :::; gm g be a generating
P set Then G
is a Gröbner basis of I if and only if for all i; j we have S(gi ; gj ) = k aijk (x1 ; :::; xn )gk
and LM (aijk gk ) LM (S(gi ; gj )).
This condition is necessary since very element of I reduces to 0 relative to
a Gröbner basis and the reduction process yields an expression with the above
property. In actual practice one mainly uses the following “weaker form”of the
criterion.
Corollary 7 Let I be an ideal. Let G = fg1 ; :::; gm g be a generating set such
that LM (gi ) 6= LM (gj ) for i 6= j.Then G is a Gröbner basis of I if and only if
for all i; j we have redG (S(gi ; gj )) = 0.
Exercises:
1. Let I be the ideal in F [x1 ; :::; x2n ] generated by fxi xn+j xj xn+i ji < jg.
Show that this is a Gröbner basis relative to Lex.
2. In this exercise we will also use Lex for our order. Let I be the ideal
generated
by e1 (x1 ; :::; xn ); e2 (x1 ; :::; xn ); :::; en (x1 ; :::; xn ) with em (x1 ; :::; xn ) =
P
xim ; ::::Use the identity
1 i1 < <im n xi1 xi2
0=
m
Q
(xm
xj ) =
j=1
m
X
( 1)m
j
em
j
j (x1 ; :::; xm )xm
j=0
to show that there is a generating set h1 = e1 ; h2 ; :::; hn = xnn for I with
LM (hj ) = xjj . (Hint: The displayed formula says that
xm
m =
m
X1
( 1)m
j
em
j
j (x1 ; :::; xm )xm :
j=0
Use ej (x1 ; :::; xm+1 ) = ej (x1 ; :::; xm ) + ej 1 (x1 ; :::; xm )xm+1 to show that the
leading term of
m
X1
( 1)m j em j (x1 ; :::; xn )xjm
j=0
is xm
m .) Prove that fh1 ; :::; hn g is a Gröbner basis for I with respect to Lex. For
this you may want to use the next exercise.
3. Let 0 6= f; g 2 F [x1 ; :::; xn ] be such that LM (f ) and LM (g) are relatively
prime. Then S(f; g) = af + bg with LM (f; g)
LM (af ) and LM (f; g)
LM (bg). (Hint: Assume LC(f ) = LC(g) = 1: Then f = LM (f ) + u and
7
g = LM (g) + v. Thus S(f; g) = LM (g)f LM (f )g = vf ug. Show that the
leading terms of vf and ug are distinct.)
We will now prove the Buchberger
Pcriterion.
We assume that each S(gi ; gj ) = aijk gk with LM (aijk gk ) S(gi ; gj ). Let
f 2 I be nonzero. Then we must show that LM
P (f ) dominates some LM (gi ).
For this it would be enough to prove that f =
bl gl with LM (bl gl ) LM (f )
fpr all l. Since then we must have some LM (bi gi ) = LM (f ) so LM (f ) dominates
LM (gi ):
P We procede
P to prove this by contradiction. Let among all expressions f =
bl gl , f =
ci gi be the one with maxfLM (ci gi )ji = 1; :::; mg = , minimal. We note that
LM (f ). We assume that
> LM (f ) and derive
a contradiction. If we relabel we may assume that LM (ci gi ) = for i
k
and LM (ci gi ) < for i > k. We write LM (gi ) = x i and we assume that
i
LC(gi ) = 1. We write
for
i + i =
PLM (ci ) = x Pand LC(ci ) = ui . Then
P
i k. We have f = i k ui x i gi + i k (ci ui x i )gi + i>k ci gi . We have
LM (ci ui x i gi ) < for i k and LM (ci gi ) < for i >P
k. We look at the
…rst term. We note that the coe¢ cient of x in that term is i k ui : This must
P
be the coe¢ cient of x in f which is 0: Thus i k ui = 0. We note that if
P
Pk 1
(z1 ; :::; zk ) 2 F k and i k zi = 0 then (z1 ; :::; zk ) = i=1 wi (ei ei+1 ) with ei
the vector with a 1 in the i-th position and 0’s everywhere else. This says that
there are elements of F , di , such that
X
X
ui x i gi =
di z(x i gi x i+1 gi+1 ):
( )
i k
i k 1
Since i + i = we see that x i gi x i+1 gi+1 = x i S(gi ; gi+1 ): for an appropriate
i We also note that LM
P(S(gi ; gi+1 )) < LM (LCM (gi ; gi+1 )): Our hypothesis
implies that S(gi ; gj ) =
aijk gk with LM (aijk gk ) LM (S(gi ; gj )). Thus we
have
X
X
ui x i gi =
di aii+1k x i gk
i k
with LM (aii+1k gk ) LM (S(gi ; gi+1 )) <
i . Now putting all of these terms
together we have an expression
X
f=
hi gi
with LM (hi gi ) < for all i. This is the desired contradiction.
Exercise. Carry out the details of the proof of (*) above.
4
Buchberger’s algorithm.
We will now give an algorithm that allows the e¤ective computation of Gröbner
bases on a computer. We start with a …xed monomial order and a generating set for an ideal ff1 ; :::; fm g in F [x1 ; :::; xn ]. As an output we will get a
8
Gröbner basis. Polynomials are partially ordered via their leading terms. After
normalization a …nite set of polynomials is totally ordered.
Here is the pseudo code:
0g.
PresentBasis = Normalze(f1 ; :::; fm ); Queue = fS(f; g)jf; g 2 Presentbasis; S(f; g) 6=
Repeat:
If Queue = ; return PresentBasis;
u a minimal element of the Queue;
v = redPresentBasis (u); Queue = Queue fug;
If v = 0; Continue;
Queue = Queue [fS(f; v)jf 2 PresentBasisg;
PresentBasis=PresentBasis[fvg;
Continue;
We must show that this pseudocode terminates. So assume that it doesn’t
terminate. Then the algorithm yields an in…nite set fv1 ; v2 ; :::g such that
LM (vi ) doesn’t dominate LM (vj ) for i 6= j. Consider
Ij = hLM (v1 ); :::; LM (vj )i
P
assume that Ij+1 = Ij then LM (vj+1 ) =
h
LM
(vi ): This implies that
i
i j
LM (vj+1 ) must be a monomial in one of the hi LM (vi ); i
j. But then we
would have LM (vj+1 ) dominates LM (vi ). This is contrary to the outcome of
the (assumed to be) failed pseudocode. Thus the ideals Ij must never stabilize
and this contradicts the Hilbert basis theorem.
In practice there are several tricks due to Buchberger to speed up the algorithm (one is related to Exercise 3 in section 3). There is another that is related
to a generaliziation of the algorithm to submodules of free modules. These improvements are only about 20%. There are no known exponential speadups.
These algorithms properly souped up are the basis of computational algebra
packages such as Mathematica, Maple, Macaulay, Singular,...
We will give algorithms based on the (assumed ) calculation of a Gröbner
basis in Mathematica code.
5
Hilbert Series
In this section we will …x a monomial order, >. We consider a homogeneous
ideal I F [x1 ; :::; xn ]. Then A = F [x1 ; :::; xn ]=I is graded by degree. Here the
m-th part of the grade is, Am , the image in A of the polynomials homogeneous
of degree m. We de…ne the formal power series
H(A; q) =
1
X
dim Am q m :
m=0
p
H(A) is called the Hilbert series of the graded algebra A. If I = I and
X = Pn 1 (I) then we will give and interpretation of the numbers dim Am later
9
in terms of sheal cohomology. In this section we will give algorithmic methods
to calculate it. We will also use the algorithm to prove some properties of this
series.
Let LM (I) be the monomial ideal associated to I corresponding to the chosen
order. Let S be a …nite set of monomial generators for LM (I). We can …nd
S by calculating a Gröbner basis of I. We will now assume that that has been
done and we have S. We look upon S as a subset of Zn 0 . If ; 2 Zn 0 we
write
if
2 Zn 0 (that is if dominates ). We write
if
doesn’t dominate . If T
Zn 0 we write
T if
for all 2 T . With
this notation in place we can begin our approach to calculation Hilbert series
Set for T Zn 0 ; B(T ) = f 2 Zn 0 j
T g. We set
Hilb(T; n) =
X
qj j :
2B(T )
From the material in section 2 we have
Lemma 8 H(A; q) = Hilb(S; n):
This says that we need only give an algorithm for the calculation of Hilb(T; n)
for T a …nite subset of Zn 0 . First if T = ; then B(T ) = Zn 0 so
Hilb(;; n) =
X
2Zn 0
qj
j
=
1
(1
q)n
;
Next we not that if 0 2 T then B(T ) = ; so Hilb(T; n) = 0.
If n = 1 and T 6= ; then T = f(a1 ); :::; (am )g if r = minfa1 ; :::; am g then
B(T ) = B(frg). Hence
X
Hilb(T; 1) =
q j = 1 + q + ::: + q r 1 :
j<r
Now assume that n > 1 and T 6= ;. Then we set m = maxf 1 j = ( 1 ; :::; n ) 2
T g. We also set for 2 Zn 0 , 0 = ( 2 ; :::; n ) and if U
Zn 0 we set U 0 =
0
f j 2 U g. We note that
I. If 2 Zn 0 and 1 m then 2 B(T ) if and only if 0 2 B(T 0 ).
If 0 i m 1 we set T (i) = f 2 T j 1 ig. We note that
II. If
2 Zn 0 1 and 0
i
m 1 then (i; ) 2 B(T ) if and only if
0
2 B(T (i) ).
We will prove the second and leave the …rst as an exercise.
Suppose that (i; ) 2 B(T ) and 2
= B(T (i)0 ). Then there is 2 T (i)0 with
. Now there exists j
i such that (j; ) 2 T and since (i; ) (j; ) we
have a contradiction. Thus if (i; ) 2 B(T ) then 2 B(T (i)0 ). If 2 B(T (i)0 )
and (i; ) 2
= B(T ) then there is a 2 T with (i; )
. But then 1 i thus
0
2 T (i) and we have a contradiction. This completes the proof of II.
This leads directly to the following (recursive) algorithm.
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Pseudocode to calculate Hilb(T; n) given a …nite subset, T , in Zn 0 . (Here
we assume n 1.)
If T = ; return
1
:
(1 q n )
If 0 2 T then return 0;
If n = 1 then
r = minf 1 j 2 T g;
return 1 + q + ::: + q r 1 ;
m = maxf 1 j 2 T g:
return
m
X1
q i Hilb(T (i)0 ; n
1) +
i=0
qm
Hilb(T 0 ; n
1 q
1).
Example. If we start with the example in section 3 where I = he1 ; :::; en i
in F [x1 ; :::; xn ]. Then we saw that the monomial ideal corresponding to Lex is
generated by
T = f(1; 0; 0; :::; 0); (0; 2; 0; :::; 0); (0; 0; 3; :::; 0); ::::; (0; 0; ::::; 0; n)g:
We will denote this as Tn . It is convenient to use the algorthm above starting
with the last variable rather than the …rst. Thus the prime notation will mean
drop the last coordinate and the T (i) notation will mean that the last coordinate
is at most i. Assume n > 1. Then we have in the algorithm m = n. If i n 1
then
T (i)0 = f(1; 0; 0; :::; 0); (0; 2; 0; :::; 0); (0; 0; 3; :::; 0); ::::; (0; 0; ::::; 0; n
1)g
whereas 0 2 T 0 . So if n > 1 the algorithm returns
n
X1
q i Hilb(T (i)0 ; n
1) = (1 + q + ::: + q n
1
)Hilb(Tn
1; n
1):
i=0
We follow the recursion and get (since Hilb(f(1); 1)) = 1).
Hilb(T1 ; 1)
nQ1
(1 + q + ::: + q j ) =
j=1
We can rewrite this as
nQ1
(1 + q + ::: + q j ):
j=1
n 1
Q
i=1 1
qi
:
q
Exercise 1. Show that the Hilbert series corresponding to F [x1 ; :::; x2n ]=I
as in Exercise 1 in section 3 is
1 + nq
:
(1 q)n+1
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We note that the algorithm implies that if T
Zn 0 then
p(q)
(1 q)n
Hilb(T; n) =
with p(q) a polynomial in q with integral coe¢ cients. In the obove example
n
Q
p(q) =
(1 q n ).
i=1
Pd
We also note that if p(q) = i=0 pi q i then p(q) is divisible by (1 q) if and
Pd
only if p(1) = i=0 pi = 0. If that is so and
P we set aj = p0 + p1 + ::: + pj then
aj = 0 for j > d 1 and p(q) = (1 q) aj q j = (1 q)h(q). If h(1) = 0 we
can do this procedure again. We therefore see that it is a simple computational
procedure to write
p(q) = (1 q)r P (q)
with P (1) 6= 0.
Proposition 9 Assume that H(A; q) has been writteh in the form
P (q)
(1 q)d
with P (1) 6= 0. Then d = dim An (I). In other words, the dimension of the
projective variety in Pn 1 that is the locus of 00 s of I is d 1.
Proof. We note that
1
q)d
(1
=
1
X
j+d 1 j
q :
d 1
j=0
Thus if P (q) = a0 + a1 q + :::: + am q m then
m
1
XX
P (q)
j + d 1 j+i
ai
q :
=
d
(1 q)
d 1
i=0 j=0
On the other hand
H(A; q) =
1
X
dim Ak q k :
k=0
This implies that
dim Ak =
m
X
ai
k
i=0
We we set
hA (t) =
m
X
ai
(t
i+d
1)(t
i=0
12
i+d
d 1
1
i + d 2)
(d 1)!
:
(t
i + 1)
This implies that if k
k
m then
dim A = h(k) = (
m
X
i=0
ai )
kd 1
+ lower degree in k.
(d 1)!
This implies that the Hilbert polynomal of the …ltered algebra Aj = A0 +:::+Aj
is
m
X
td
(
ai ) + lower degree in t:
d!
i=0
Pm
We note that since i=0 ai 2 Z f0g that d is indeed the degree of the Hilbert
polynomial.
We will call hA (t) the graded (projective) Hilbert polynomial. It has the
property that
dim Ak = hA (k) for k >> 0:
Exercise 2. Show that if I is as in exercise 1 then dim A2n (I) = n + 1.
Calculate the graded Hilbert polynomial of A = F [x1 ; :::; x2n ]=I.
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